Difference between revisions of "2019 AMC 10B Problems/Problem 11"

Revision as of 23:27, 8 June 2022

Problem

Two jars each contain the same number of marbles, and every marble is either blue or green. In Jar $1$ the ratio of blue to green marbles is $9:1$ , and the ratio of blue to green marbles in Jar $2$ is $8:1$ . There are $95$ green marbles in all. How many more blue marbles are in Jar $1$ than in Jar $2$ ?

$\textbf{(A) } 5\qquad\textbf{(B) } 10 \qquad\textbf{(C) }25 \qquad\textbf{(D) } 45 \qquad \textbf{(E) } 50$

Solution 1

Call the number of marbles in each jar $x$ (because the problem specifies that they each contain the same number). Thus, $\frac{x}{10}$ is the number of green marbles in Jar $1$ , and $\frac{x}{9}$ is the number of green marbles in Jar $2$ . Since $\frac{x}{9}+\frac{x}{10}=\frac{19x}{90}$ , we have $\frac{19x}{90}=95$ , so there are $x=450$ marbles in each jar.

Because $\frac{9x}{10}$ is the number of blue marbles in Jar $1$ , and $\frac{8x}{9}$ is the number of blue marbles in Jar $2$ , there are $\frac{9x}{10}-\frac{8x}{9}=\frac{x}{90} = 5$ more marbles in Jar $1$ than Jar $2$ . This means the answer is $\boxed{\textbf{(A) } 5}$ .

Solution 2 (Completely Solve)

Let $b_1$ , $g_1$ , $b_2$ , $g_2$ , represent the amount of blue marbles in jar 1, the amount of green marbles in jar 1, the the amount of blue marbles in jar 2, and the amount of green marbles in jar 2, respectively. We now have the equations, $\frac{b_1}{g_1} = \frac{9}{1}$ , $\frac{b_2}{g_2} = \frac{8}{1}$ , $g_1 + g_2 =95$ , and $b_1 + g_1 = b_2 + g_2$ . Since $b_1 = 9g_1$ and $b_2 = 8g_2$ , we substitute that in to obtain $10g_1 = 9g_2$ . Coupled with our third equation, we find that $g_1 = 45$ , and that $g_2 = 50$ . We now use this information to find $b_1 = 405$ and $b_2 = 400$ .

Therefore, $b_1 - b_2 = 5$ so our answer is $\boxed{\textbf{(A) } 5}$ . ~Binderclips1

~LaTeX fixed by Starshooter11 ~Typo fixed by Little

Solution 3

Call the number of green marbles in Jar 1 $x$ , and the number of green marbles in Jar 2 $y$ . Realize that this is the Diophantine equation $10x+9y=95$ where the only solution we care about is the integer one. Immediately realize that since $10x$ will end with a $0$ no matter what, $9y$ must end in a $5$ . The only $y$ that works is $y=5$ which leads to $x=5$ and we find that there are $45$ blue marbles in Jar 1, and $40$ in Jar 2. The difference is $45-40=5$ . ~JH. L

Video Solution

https://youtu.be/mXvetCMMzpU

~IceMatrix

@@ Line 26: / Line 26: @@
 ~LaTeX fixed by Starshooter11
 ~Typo fixed by Little
+==Solution 3==
+Call the number of green marbles in Jar 1 <math>x</math>, and the number of green marbles in Jar 2 <math>y</math>.
+Realize that this is the Diophantine equation <math>10x+9y=95</math> where the only solution we care about is the integer one. Immediately realize that since <math>10x</math> will end with a <math>0</math> no matter what, <math>9y</math> must end in a <math>5</math>. The only <math>y</math> that works is <math>y=5</math> which leads to <math>x=5</math> and we find that there are <math>45</math> blue marbles in Jar 1, and <math>40</math> in Jar 2. The difference is <math>45-40=5</math>.
+~JH. L
 ==Video Solution==

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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